ANSWER :
The answer is B.
[tex]x=\frac{c}{1+\frac{\tan\beta}{\tan\theta}}[/tex]
EXPLANATION :
Note that the tangent of an angle is :
[tex]\tan\theta=\frac{opposite}{adjacent}[/tex]
For angle B :
[tex]\begin{gathered} \tan\theta=\frac{h}{c-x} \\ h=(c-x)\tan\theta \end{gathered}[/tex]
For angle A :
[tex]\begin{gathered} \tan\beta=\frac{h}{x} \\ h=x\tan\beta \end{gathered}[/tex]
Equate both functions in h :
[tex]\begin{gathered} h=h \\ (c-x)\tan\theta=x\tan\beta \\ c\tan\theta-x\tan\theta=x\tan\beta \\ c\tan\theta=x\tan\theta+x\tan\beta \\ c\tan\theta=x(\tan\theta+\tan\beta) \\ x=\frac{c\tan\theta}{\tan\theta+\tan\beta} \\ \text{ Multiply the numerator and denominator by 1 over tan \lparen theta\rparen :} \\ x=\frac{c\tan\theta\times\frac{1}{\tan\theta}}{(\tan\theta+\tan\beta)\times\frac{1}{\tan\theta}} \\ x=\frac{c}{1+\frac{\tan\beta}{\tan\theta}} \end{gathered}[/tex]
